3.1310 \(\int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=301 \[ -\frac {d^2 \left (2 a c d-b \left (c^2 (2-m)-d^2 m\right )\right ) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right )}{f (m+1) \left (c^2+d^2\right )^2 (b c-a d)^2}+\frac {d^2 (a+b \tan (e+f x))^{m+1}}{f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))}+\frac {(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (c-i d)^2}-\frac {(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a) (c+i d)^2} \]

[Out]

1/2*hypergeom([1, 1+m],[2+m],(a+b*tan(f*x+e))/(a-I*b))*(a+b*tan(f*x+e))^(1+m)/(I*a+b)/(c-I*d)^2/f/(1+m)-1/2*hy
pergeom([1, 1+m],[2+m],(a+b*tan(f*x+e))/(a+I*b))*(a+b*tan(f*x+e))^(1+m)/(I*a-b)/(c+I*d)^2/f/(1+m)-d^2*(2*a*c*d
-b*(c^2*(2-m)-d^2*m))*hypergeom([1, 1+m],[2+m],-d*(a+b*tan(f*x+e))/(-a*d+b*c))*(a+b*tan(f*x+e))^(1+m)/(-a*d+b*
c)^2/(c^2+d^2)^2/f/(1+m)+d^2*(a+b*tan(f*x+e))^(1+m)/(-a*d+b*c)/(c^2+d^2)/f/(c+d*tan(f*x+e))

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Rubi [A]  time = 0.80, antiderivative size = 299, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3569, 3653, 3539, 3537, 68, 3634} \[ -\frac {d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right ) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right )}{f (m+1) \left (c^2+d^2\right )^2 (b c-a d)^2}+\frac {d^2 (a+b \tan (e+f x))^{m+1}}{f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))}+\frac {(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (c-i d)^2}-\frac {(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a) (c+i d)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^m/(c + d*Tan[e + f*x])^2,x]

[Out]

(Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a - I*b)]*(a + b*Tan[e + f*x])^(1 + m))/(2*(I*a + b)
*(c - I*d)^2*f*(1 + m)) - (Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a + I*b)]*(a + b*Tan[e + f
*x])^(1 + m))/(2*(I*a - b)*(c + I*d)^2*f*(1 + m)) - (d^2*(2*a*c*d - b*c^2*(2 - m) + b*d^2*m)*Hypergeometric2F1
[1, 1 + m, 2 + m, -((d*(a + b*Tan[e + f*x]))/(b*c - a*d))]*(a + b*Tan[e + f*x])^(1 + m))/((b*c - a*d)^2*(c^2 +
d^2)^2*f*(1 + m)) + (d^2*(a + b*Tan[e + f*x])^(1 + m))/((b*c - a*d)*(c^2 + d^2)*f*(c + d*Tan[e + f*x]))

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3653

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2))/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rubi steps

\begin {align*} \int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^2} \, dx &=\frac {d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac {\int \frac {(a+b \tan (e+f x))^m \left (-a c d+b \left (c^2-d^2 m\right )-d (b c-a d) \tan (e+f x)-b d^2 m \tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx}{(b c-a d) \left (c^2+d^2\right )}\\ &=\frac {d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac {\int (a+b \tan (e+f x))^m \left ((b c-a d) \left (c^2-d^2\right )-2 c d (b c-a d) \tan (e+f x)\right ) \, dx}{(b c-a d) \left (c^2+d^2\right )^2}-\frac {\left (d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right )\right ) \int \frac {(a+b \tan (e+f x))^m \left (1+\tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx}{(b c-a d) \left (c^2+d^2\right )^2}\\ &=\frac {d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac {\int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx}{2 (c-i d)^2}+\frac {\int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx}{2 (c+i d)^2}-\frac {\left (d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right )\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^m}{c+d x} \, dx,x,\tan (e+f x)\right )}{(b c-a d) \left (c^2+d^2\right )^2 f}\\ &=-\frac {d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right ) \, _2F_1\left (1,1+m;2+m;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m}}{(b c-a d)^2 \left (c^2+d^2\right )^2 f (1+m)}+\frac {d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac {i \operatorname {Subst}\left (\int \frac {(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 (c-i d)^2 f}-\frac {i \operatorname {Subst}\left (\int \frac {(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d)^2 f}\\ &=\frac {\, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (i a+b) (c-i d)^2 f (1+m)}-\frac {\, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (i a-b) (c+i d)^2 f (1+m)}-\frac {d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right ) \, _2F_1\left (1,1+m;2+m;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m}}{(b c-a d)^2 \left (c^2+d^2\right )^2 f (1+m)}+\frac {d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 4.36, size = 266, normalized size = 0.88 \[ \frac {(a+b \tan (e+f x))^{m+1} \left (-\frac {2 d^2 \left (2 a c d+b c^2 (m-2)+b d^2 m\right ) \, _2F_1\left (1,m+1;m+2;\frac {d (a+b \tan (e+f x))}{a d-b c}\right )}{(m+1) \left (c^2+d^2\right ) (a d-b c)}-\frac {i \left (\frac {(c-i d)^2 (b c-a d) \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a+i b}\right )}{a+i b}+\frac {(c+i d)^2 (a d-b c) \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a-i b}\right )}{a-i b}\right )}{(m+1) \left (c^2+d^2\right )}-\frac {2 d^2}{c+d \tan (e+f x)}\right )}{2 f \left (c^2+d^2\right ) (a d-b c)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^m/(c + d*Tan[e + f*x])^2,x]

[Out]

((a + b*Tan[e + f*x])^(1 + m)*(((-I)*(((c + I*d)^2*(-(b*c) + a*d)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Ta
n[e + f*x])/(a - I*b)])/(a - I*b) + ((c - I*d)^2*(b*c - a*d)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e +
f*x])/(a + I*b)])/(a + I*b)))/((c^2 + d^2)*(1 + m)) - (2*d^2*(2*a*c*d + b*c^2*(-2 + m) + b*d^2*m)*Hypergeomet
ric2F1[1, 1 + m, 2 + m, (d*(a + b*Tan[e + f*x]))/(-(b*c) + a*d)])/((-(b*c) + a*d)*(c^2 + d^2)*(1 + m)) - (2*d^
2)/(c + d*Tan[e + f*x])))/(2*(-(b*c) + a*d)*(c^2 + d^2)*f)

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fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{m}}{d^{2} \tan \left (f x + e\right )^{2} + 2 \, c d \tan \left (f x + e\right ) + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^m/(c+d*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e) + a)^m/(d^2*tan(f*x + e)^2 + 2*c*d*tan(f*x + e) + c^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^m/(c+d*tan(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^m/(d*tan(f*x + e) + c)^2, x)

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maple [F]  time = 1.76, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \tan \left (f x +e \right )\right )^{m}}{\left (c +d \tan \left (f x +e \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^m/(c+d*tan(f*x+e))^2,x)

[Out]

int((a+b*tan(f*x+e))^m/(c+d*tan(f*x+e))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^m/(c+d*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)^m/(d*tan(f*x + e) + c)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x))^m/(c + d*tan(e + f*x))^2,x)

[Out]

int((a + b*tan(e + f*x))^m/(c + d*tan(e + f*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{m}}{\left (c + d \tan {\left (e + f x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))**m/(c+d*tan(f*x+e))**2,x)

[Out]

Integral((a + b*tan(e + f*x))**m/(c + d*tan(e + f*x))**2, x)

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